Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {
    private static final int MOD = (int) (1e9+7);

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        double[] arr = new double[n+1];
        arr[0] = 1;
        for(int i =0;i<n ; ++i){
            double p = scanner.nextDouble();
            for(int r = i+1 ; r>0;--r){
                arr[r] = arr[r]*(1-p) + arr[r-1]*p;
            }
            arr[0] = arr[0]*(1-p);
        }
        double ans = 0;
        for(int i = 0 ; i<=n ; ++i){
            if(i>n-i) ans+=arr[i];
        }
        System.out.println(ans);
    }
}

























/*
*         double[] arr = new double[n+1];
        arr[0] = 1;
        for(int i=0;i<n ; ++i){
            double p = scanner.nextDouble();
            for(int r = i+1 ; r>=1 ; --r){
                arr[r] = arr[r]*(1-p) + arr[r-1]*p;
            }
            arr[0] = arr[0]*(1-p);
        }
        double ans=0;
        for(int i = 0 ; i <=n ; ++i){
            if(i>n-i) ans+=arr[i];
        }
        System.out.println(ans);
*/